Blue numbers wow

Number Theory Level 2

A positive integer $n$ is called blue, if there exist three (not necessarily distinct) numbers $>1$ ( $a, b, c,$ ), such that $n=abc$ .

Does there exist 1000000 consecutive postive integers, such that each of them is blue?

No, there doesn’t exist Yes, there exist

1 solution

Patrick Corn
Nov 24, 2017

Let $k=1000000.$ There is a sequence of $k$ consecutive composite numbers for any $k$ : e.g. $(k+1)! + 2, (k+1)! + 3, \ldots, (k+1)!+(k+1).$

Now let $m = (k+1)!+2$ and let $n = (k+1)!+(k+1),$ so the sequence of $k$ consecutive integers from $m$ to $n$ consists of composite numbers. Then $n!+m, n!+(m+1), \ldots, n!+n$ is the sequence we want.

But being composite isn't enough. For example, any number of the form $pq$ , where $p$ and $q$ are primes, is composite but not blue.

Jon Haussmann - 3 years, 6 months ago

Right, but my sequence is blue: $n!+d = de$ for some $e > 1,$ and $d$ is composite, so $de$ is blue.

Patrick Corn - 3 years, 6 months ago

Ok, thanks for clarifying.

Jon Haussmann - 3 years, 6 months ago

Blue numbers wow

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